Team Review

• How did everyone feel about the problem set?

• Any questions?

• Discuss with Team

Rules for Derivatives Continued

\[ \large \frac{d}{dx}=[f(x)*g(x)]=f(x)g'(x)+g(x)f'(x) \]

• Breakdown multiplication into two separate functions

• Then take the first function times the derivative of the second, and add the second function times the derivative of the first.

Product Rule Example

\[ \begin{align} y&=2x(x^2+3x) \\ y&=\overbrace{2x}^{f(x)}\overbrace{(x^2+3x)}^{g(x)} &\text{Break into two functions} \\ \frac{dy}{dx}&=\overbrace{2x}^{f(x)}\overbrace{(2x+3)}^{g'(x)}+\overbrace{2}^{f'(x)}\overbrace{(x^2+3x)}^{g(x)} &\text{Apply Power Rule} \\ \frac{dy}{dx}&=6x^2+12x &\text{Algebra clean up} \end{align} \]

Rule for Derivatives Continued

\[ \large \frac{d}{dx}\left [\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} \]

• Same idea as product rule, but order matters

• The denominator has to be \(g(x)\)

Quotient Rule Example

\[ \begin{align} y&=\frac{x^2}{(x+1)} \\[3px] y&=\class{fragment}{{}\frac{\overbrace{x^2}^{f(x)}}{\underbrace{(x+1)}_{g(x)}}} &\text{Identify Functions}\\[3px] \frac{dy}{dx}&=\class{fragment}{{}\frac{\overbrace{(x+1)}^{g(x)}\overbrace{2x}^{f'(x)}-\overbrace{x^2}^{f(x)}\overbrace{(1)}^{g'(x)}}{\underbrace{(x+1)^2}_{[g(x)]^2}}} &\text{Quotient Rule}\\[3px] \frac{dy}{dx}&=\class{fragment}{{}\frac{x^2+2x}{(x+1)^2}} \end{align} \]

Note: Algebraically solving Quotient Rules can be very challenging. Try to reduce it down to a usable form.

Rule of Derivatives Continued

\[ \large \frac{d}{dx}[f(g(x))]=f'(g(x))g'(x) \]

• Take the derivative of the outermost function then multiply by the derivative of the innermost

• Probably the most important rule for advanced calculus

Chain Rule Example

• Use a trick called u-substitution

• Label the inner function as u

• Take the derivative with respect to u

\[ \begin{align} y&\class{fragment}{{}=(x^3-2)^2}\\[3px] u&=\class{fragment}{{}x^3-2}\\[3px] y&\class{fragment}{{}=u^2} \\[3px] \frac{dy}{dx}&=\class{fragment}{{}\frac{dy}{du}\cdot\frac{du}{dx}} \end{align} \]

Chain Rule Example Continued

\[ \begin{align} \frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\[3px] \frac{dy}{du}&=\class{fragment}{{}2u} &\text{Reminder: }y=u^2 \\[3px] \frac{du}{dx}&=\class{fragment}{{}3x^2} &\text{Reminder: }u=(x^3-2) \\[3px] \frac{dy}{dx}&=\class{fragment}{{} 2u\cdot3x^2} \\[3px] \frac{dy}{dx}&=\class{fragment}{{} 2(x^3-2)(3x^2)} \end{align} \]

An aside on exponents

Base \(e\) exponents are unique and useful in many math applications

Originally derived to solve compounding interest rate problems in 1683 by Bernoulli

Derivatives of Exponents

\[ \large \frac{d}{dx}e^x=e^x \]

Derivatives of Exponents

Higher Order Derivatives

• We can continue to take derivatives of derivatives

• The second derivative becomes the rate of change of the rate of change (Think acceleration)

• Notation:

  • \(f''(x)\) or \(\frac{d^2y}{dx^2}\): 2nd Derivative

  • \(f^{(n)}(x)\) or \(\frac{d^ny}{dx^n}\) nth derivative

Identify which rule works best for each function

\[ \begin{align} \text{A) }y=e^{\sqrt{3x+10}} & &\text{B) }g(y)=\frac{(8y+2)}{(4y+3)^3} & & \text{C) }p(b)=18b\cdot e^{2b} \end{align} \]

1. Find the first derivative

\[ f(x)=(2x-x^2)^3 \]

2. Find the first derivative

\[ f(x)=\frac{2}{x^2+1} \]

3. Find the third derivative

\[ f(x)=x^2(2x+1) \]

Exponential Functions in the world

Infectious diseases

Human population

Communication

Growth of Renewable Energy

Logistic Growth

Resource limits do exist in the world

\[ \begin{align} \frac{dN}{dt}&=rN\left(\frac{K-N}{K}\right) \\ \text{Where }N=\text{population, }r=\text{growth rate, }&t=\text{time, and }K=\text{Carrying Capacity} \end{align} \]

Extremely important in Fisheries and Biological Conservation

Application of derivatives allows us to find crucial information to solve environmental problems

1. Find instantaneous rate of change at any point

2. Find when functions are increasing or decreasing

3. Find the Optimal values of functions

   • Optimal values are key to designing effective policies

How quickly is the human population rising right now?

Dochy et al. (1995) used the following function to describe the increase in human population over time:

\[ x(t)=\frac{1}{c-\beta t} \]

Where \(c=9.833*10^{-3}\), \(\beta=4.849*10^{-6}\), and \(x\) is the total human population in millions at year \(t\).

• How quickly was the human population increasing in 1450?

• What should be the population growth rate in 2024 according to this model?

Begin by taking the derivative

First, think of which rules we would use for \(x(t)=\frac{1}{c-\beta t}\)

Begin by taking the derivative

\[ \begin{align} x(t)&=\frac{1}{c-\beta t} \\ x(t)&=(c-\beta t)^{-1} &\text{Rearrange} \\ \frac{dx}{dt}&=\beta(c-\beta t)^{-2} &\text{Chain and Power Rule}\\ x'(1450)&=.61 \text{ million per year} &x'(2024)=13,980 \text{ million per year} \end{align} \]

The model isn’t wrong, just not predictive out of sample

Derivatives help us understand the behavior of functions and also people

• Useful in designing predictive theory for new policies

• Personal Example: Will a fishery insurance program incentivize overfishing?

\[ \begin{align} \begin{bmatrix}\frac{\partial K}{\partial \gamma} \\ \frac{\partial L}{\partial \gamma}\end{bmatrix} =-\begin{bmatrix} \frac{\partial U}{\partial K \partial K} & \frac{\partial U}{\partial K \partial L} \\ \frac{\partial U}{\partial L \partial K} & \frac{\partial U}{\partial L \partial L}\end{bmatrix}^{-1}\begin{bmatrix}\frac{\partial U}{\partial K \partial \gamma}\\ \frac{\partial U}{\partial L \partial \gamma}\end{bmatrix} \end{align} \]

• What conditions lead to:

\[ \begin{align} \begin{bmatrix}\frac{\partial K}{\partial \gamma} \\ \frac{\partial L}{\partial \gamma}\end{bmatrix}\lesseqgtr 0 \end{align} \]

Let’s start a little simpler

Definition of increasing and decreasing Intervals of a Function

• For the function \(f(x)\) (differentiable on the interval \((a,b)\))

  • If \(f'(x)>0\) for all \(x\) values between \(a\) and \(b\), this interval is increasing

  • If \(f'(x)<0\) for all \(x\) values betwen \(a\) and \(b\), this interval is said to be decreasing

  • If \(f'(x)=0\) for all values between \(a\) and \(b\), then the function is constant within the interval

Graphical demonstration of increasing vs decreasing

Optimization problems find relative extremum (aka max or min) of functions

1. Find First Order Conditions:

• Relative extremum exist at all points on function where \(f'(x)=0\)

Graphical Example of Optimization

Solve Analytically

First Order Conditions:

\[ \begin{align} f(x)&=x^3-12x+2 \\ f'(x)&=3x^2-12 \\ x^*&=-2 & x^*=2 \end{align} \]

1. How many local min and maximums can you find on this graph?

2. Solve for all optimum. Identify which are minima and which are maxima.

\[ f(x)=2x^3-6x^2+3x+15 \]

3. Your team has been hired to help the city of Santa Barbara to help eliminate a nasty pollutant in our streams. You are choosing between two policies. Policy A provides \(\sqrt{10x+4}\) benefits where \(x\) is the dollar spent on interventions. Policy B provides \(\sqrt{20x+4}\) benefits for every dollar spent. However, each policy has a cost associated with each. Policy A has a constant linear cost at \(2x\) and Policy B has a constant linear cost at \(3x\).

1. Find the optimal amount of interventions dollars to spend for each policy. Consider both costs and benefits. [hint: Remember \(\sqrt{}\) is just \(x^{\frac{1}{2}}\). You can use exponent rules to help solve]

2. Which policy provides the most benefits?

3. Which policy is cheaper?

4. Which policy should the city of Santa Barbara pursue?